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We can write it as nums1[i] - nums2[i] > nums2[j] - nums1[j] instead of nums1[i] + nums1[j] > nums2[i] + nums2[j].
Store nums1[idx] - nums2[idx] in a data structure.
Store nums2[idx] - nums1[idx] in a different data structure.
For each integer in the first data structure, count the number of the strictly smaller integers in the second data structure with a larger index in the original array.
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Yes, after sorting the difference array, binary search can find the first index that forms a valid pair for each element. This approach still runs in O(n log n) time due to sorting and repeated searches.
Yes, this type of problem is common in coding interviews because it tests array transformation, sorting, and pair-counting techniques. It also checks whether candidates can optimize a brute-force O(n^2) solution using two pointers or binary search.
The optimal approach transforms the problem using a difference array where diff[i] = nums1[i] - nums2[i]. After sorting this array, a two-pointer technique efficiently counts pairs where diff[i] + diff[j] > 0, reducing the complexity to O(n log n).
The difference array simplifies the condition involving two arrays into a single-array pair sum problem. Instead of comparing sums from both arrays separately, we only check whether the sum of two differences is positive.